Solve 48x^2y^2-3y^4

$$48 x ^ { 2 } y ^ { 2 } - 3 y ^ { 4 }$$

Factor

$$3\left(4x-y\right)\left(4x+y\right)y^{2}$$

Solution Steps

Factor out 3.

$$3\left(16x^{2}y^{2}-y^{4}\right)$$

Consider 16x^{2}y^{2}-y^{4}. Factor out y^{2}.

$$y^{2}\left(16x^{2}-y^{2}\right)$$

Consider 16x^{2}-y^{2}. Rewrite 16x^{2}-y^{2} as \left(4x\right)^{2}-y^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(4x-y\right)\left(4x+y\right)$$

Rewrite the complete factored expression.

$$3y^{2}\left(4x-y\right)\left(4x+y\right)$$

Evaluate

$$48\left(xy\right)^{2}-3y^{4}$$